Question 29563
I'm not quite sure about what the first problem is asking. Try rewriting it with the parenthesis or symbols to clarify it.  I'll try to answer the second one instead.<br>Starting with {{{(3x-8)^(1/2)=5}}} You can't just drop the parenthesis... they have meaning. To rewrite without them as {{{3x-8^(1/2)=5}}} would change the problem.  Instead, as we solve virtually every algebra equation, we have to do the same thing to both sides. Since we're first concerned with the 1/2 exponent outside the parentesis on the right, we can get rid of it by squaring both sides:
{{{((3x-8)^(1/2))^2=5^2}}}The rule here being that by to eliminate the exponent (1/2) you want to raise the exponent to the recipical (2/1=2). When you take an exponent and raise it to another exponent, you multiply the two together. In brief: {{{(x^a)^b=x^(a*b)}}} So in our example 1/2*2=1 so we end up with:
{{{3x-8=25}}}Now this should look like something familiar and you can solve by adding 8 to both sides:
{{{3x=33}}}Divide by 3:
{{{x=11}}}
Double checking:
{{{(3x-8)^(1/2)=5}}}
{{{(3(11)-8)^(1/2)=5}}}
{{{(33-8)^(1/2)=5}}}
{{{(25)^(1/2)=5}}}Check.